(Photo from Hershey Park Website)
The tidal force is basically a log that is carried up a huge lift then goes down a big hill and enters the water below creating a HUGE splash. Since I've lost all my info on this page three types I am just going to copy and paste my entire document.
Basic Info: Height of lift = 30.5 meters
full boat is 4082 kg, empty is 2722 kg.
g = 9.80 m/s^2
Potential Energy at top = FULL = 1220109.8, EMPTY = 813605.8
When AT GROUND, velocity = 24.66
More info: Water gets shallower at about 10 meters. This will speed up the boat. Although the actual change in depth is unknown, we will say that it was half as deep at the end than the beginning. Each Log was about half full or 3402 kg
DATA COLLECTED AT PARK
Angle of Lift – 27 degrees
Time of Lift – 47.25 seconds
Distance of lift (approximation) – 42.98 m
Time from entering water to reach 18.29 m – 2.62, 2.53, 2.78, 2.48
Speed of Cart before drop = 2.25 m/s
Time of Splash – 3.45, 3.41, 3.55, 3.39
Distance of Splash – 40 yards, 42 yards, 37 yards, 41 yards
Time of Drop – 4.87, 4.71, 4.89, 4.89
ANALYSIS
Work = F * d
Force Horizontal = mg, Mass was 3402 * 9.8 = 33339.6 * 42.98 m = 1370.4 kJ= Work to go up hill.
Velocity entering the water was 24.66 m/s so it would take about .742 s to reach the designated marker
It took 2.6 seconds to reach the end point. So the acceleration can be found using ∆x = 24.66*2.6 + .5 * a * 2.6^2 -> -13.55 = a
Velocity at 10 meters v^2 = 24.66^2 + 2*-13.55*10 = 337 = v^2 -> v = 18.35 at ten meters
Velocity at end v^2 = 24.66^2 + 2*-13.55*18.29 = 112.46 = v^2 -> v = 10.60
Then it can be found how long it took to get to that point. 10 = (24.66 + 18.35 / 2) * ∆t -> t = .465 seconds.
With that, it took 2.13 to travel the last eight, so the deceleration at that point is much greater.
To find launch angle. Wave’s initial speed cos ø = X Velocity
Wave went 36.57 meters and took 3.45 sec to land so the X velocity is 10.6 m/s
10.6 = 24.45 cos ø -> launch angle = 64.3
GRAPHS AND TABLES
Velocity vs Distance Graph. Y = Velocity X = Distance from entry into water
Position vs Time Graph. Y = Distance from entry into water, X = Time
Rough sketch of Wave (Note that the top of the wave is indicated by the line.)
Table indicating points on the graphs given.
Sorry for the sloppy layout, I just actually wanted this done before it gets wiped out again.
The tidal force is basically a log that is carried up a huge lift then goes down a big hill and enters the water below creating a HUGE splash. Since I've lost all my info on this page three types I am just going to copy and paste my entire document.
Basic Info: Height of lift = 30.5 meters
full boat is 4082 kg, empty is 2722 kg.
g = 9.80 m/s^2
Potential Energy at top = FULL = 1220109.8, EMPTY = 813605.8
When AT GROUND, velocity = 24.66
More info: Water gets shallower at about 10 meters. This will speed up the boat. Although the actual change in depth is unknown, we will say that it was half as deep at the end than the beginning.
Each Log was about half full or 3402 kg
DATA COLLECTED AT PARK
Angle of Lift – 27 degrees
Time of Lift – 47.25 seconds
Distance of lift (approximation) – 42.98 m
Time from entering water to reach 18.29 m – 2.62, 2.53, 2.78, 2.48
Speed of Cart before drop = 2.25 m/s
Time of Splash – 3.45, 3.41, 3.55, 3.39
Distance of Splash – 40 yards, 42 yards, 37 yards, 41 yards
Time of Drop – 4.87, 4.71, 4.89, 4.89
ANALYSIS
Work = F * d
Force Horizontal = mg, Mass was 3402 * 9.8 = 33339.6 * 42.98 m = 1370.4 kJ = Work to go up hill.
Velocity entering the water was 24.66 m/s so it would take about .742 s to reach the designated marker
It took 2.6 seconds to reach the end point. So the acceleration can be found using ∆x = 24.66*2.6 + .5 * a * 2.6^2 -> -13.55 = a
Velocity at 10 meters v^2 = 24.66^2 + 2*-13.55*10 = 337 = v^2 -> v = 18.35 at ten meters
Velocity at end v^2 = 24.66^2 + 2*-13.55*18.29 = 112.46 = v^2 -> v = 10.60
Then it can be found how long it took to get to that point. 10 = (24.66 + 18.35 / 2) * ∆t -> t = .465 seconds.
With that, it took 2.13 to travel the last eight, so the deceleration at that point is much greater.
To find launch angle. Wave’s initial speed cos ø = X Velocity
Wave went 36.57 meters and took 3.45 sec to land so the X velocity is 10.6 m/s
10.6 = 24.45 cos ø -> launch angle = 64.3
GRAPHS AND TABLES
Velocity vs Distance Graph. Y = Velocity X = Distance from entry into water
Position vs Time Graph. Y = Distance from entry into water, X = Time
Rough sketch of Wave (Note that the top of the wave is indicated by the line.)
Table indicating points on the graphs given.
Sorry for the sloppy layout, I just actually wanted this done before it gets wiped out again.